2 Preliminaries

Although it seems difficult to find a positive integer n so that $n^2+4$ is square-full, there are, in fact, infinitely many such values of n. Let $n=2d$ be an even number and rewrite $n^2+4=4(d^2+1)$ . There are infinitely many solutions $(d,k)\in \mathbb Z^2$ such that

$$ \begin{align*}4(d^2+1)=8k^2\Longleftrightarrow d^2-2k^2=-1,\end{align*} $$

since it is a negative Pell’s equation with $(1,1)$ as a base solution. Since $8k^2$ is always square-full, we produce infinitely many square-full numbers of the form $n^2+4$ . A polynomial $f(x)$ of degree at least three with no repeated roots is unlikely to take square-full values infinitely often. This is thanks to the result of Granville as a consequence of the $abc$ conjecture (see [Reference Granville10, Theorem 7] and the discussion around it). Our example $f(x)=x^2+4$ shows that the number of $n\leqslant N$ for which $f(n)$ is square-full is $\gg \log N$ . We may ask, as mentioned by Granville, whether

(2.1) $$ \begin{align} S^{{\mathstrut\hspace{0.05em}\blacksquare}}_f(N)\sim c_f\log N \end{align} $$

for some constant $c_f$ depending on the coefficients of $f(x)$ . In particular, this might hold for any admissible quadratic polynomial $f(x)$ with no fixed prime p such that $p\,\|\,f(n)$ for all sufficiently large n. In Proposition 3.7, we shall see that assuming the $abc$ conjecture, the left-hand side of (2.1) is bounded by $C_{\varepsilon ,f} N^\varepsilon $ for some constant $C_{\varepsilon , f}>0$ . By the Erdős–Szekeres theorem, the probability of n in each dyadic range $[{N}/{2},N]$ to be square-full is of order $1/\sqrt {N/2}$ . Thus, we expect that the probability of both n and $n+1$ in the same range to be square-full is of order $1/(N/2)$ . Thus, it is reasonable to expect the number of consecutive square-full integers up to N to be of order

$$ \begin{align*}\dfrac{N}{2}\cdot\dfrac{1}{N/2}+\dfrac{N}{4}\cdot\dfrac{1}{N/4}+\cdots \ll\log N,\end{align*} $$

where the sum ranges through all dyadic intervals not exceeding N. This gives a simple heuristic argument that supports (2.1).

For a fixed polynomial $f(x)$ , it is convenient to study the expression

(2.2) $$ \begin{align} S^{{\mathstrut\hspace{0.05em}\blacksquare}}_f(2N)-S^{{\mathstrut\hspace{0.05em}\blacksquare}}_f(N) =\sum_{\substack{N<n\leqslant 2N\\ e^2d^3=f(n)}} \mu^2(d); \end{align} $$

an upper bound gives a bound for $S^{{\mathstrut \hspace {0.05em}\blacksquare }}_f(N)$ itself by a dyadic sum, at the cost of a logarithmic factor. By defining

$$ \begin{align*}\mathcal{M}(E,D) := \#\{(e,d,n)\in\mathbb N^3: N<n\leqslant 2N, E<e\leqslant 2E, D<d\leqslant 2D, e^2d^3=f(n)\},\end{align*} $$

the right-hand side of (2.2) is at most

$$ \begin{align*} \sum_{\ell\ll \log N} \sum_{\substack{{N}/{2^{\ell}}<E\leqslant {N}/{2^{\ell-1}}\\E^2D^3\asymp_f N^2}}\mathcal{M}(E,D)\ll (\log N)\sup_{\substack{E\leqslant N\\ E^2D^3\asymp_f N^2}}\mathcal{M}(E,D), \end{align*} $$

where the sums are dyadic sums over E. From this, the problem is reduced to obtaining the bound for $\mathcal {M}(E,D)$ in the entire range. Estermann [Reference Estermann9] showed that the contribution to $\mathcal {M}(E,D)$ for fixed $d\in [D,2D]\subset [1,N]$ is $O(\log N)$ , so we have $\mathcal {M}(E,D)\ll D\log N$ .

It seems difficult to extend Heath-Brown’s determinant method to a polynomial $f(x)=x^2+h$ with nonsquare h. For general quadratic polynomials, we use a different approach in Section 3, employing the bound on counting integral points on the Mordell curve $y^2=x^3+D$ for a nonzero integer D.

In Section 4, we prove our first main theorem on $f(x)=x^2+\alpha ^2$ . We proceed in parallel to Heath-Brown’s work [Reference Heath-Brown13] on square-free values of $n^2+1$ , by refining the bound of $\mathcal {M}(E,D)$ to $N^\eta \sqrt {M}\min (\sqrt {E},\sqrt {D})$ , with a new parameter M. This leads to the construction of the vanishing equation and addresses our challenges mentioned earlier. Finally, in Section 5, we provide some comments on the generalisation to higher-degree polynomials and the reduction of $\varpi $ in both main theorems.

3 Integral points on Mordell curves and the proof of Theorem 1.2

3.1 Proof of Theorem 1.2 and counting integral points on Mordell curves

First, we state the conjecture on counting integral solutions on Mordell curves.

Conjecture 3.1 (Mordell curve conjecture).

For a fixed integer $D\not =0$ and an arbitrarily small $\varepsilon>0$ , we have $\#\{(x,y)\in \mathbb Z^2: y^2=x^3+D\} \ll _\varepsilon |D|^{\varepsilon }$ .

Using the Helfgott–Venkatesh mechanism [Reference Helfgott and Venkatesh14] with the Ellenberg–Venkatesh result [Reference Ellenberg and Venkatesh7], we prove the following proposition.

Proposition 3.2. Let $\varepsilon>0$ be given and let $\varpi _0={2\beta }/{3\log 3}\approx 0.1688$ be as in (1.2). For a fixed nonzero integer D, the number of integer solutions on the Mordell curve $y^2=x^3+D$ is at most $O_\varepsilon (|D|^{\varpi _0+\varepsilon })$ .

Proof. Substitute the bound for the $3$ -torsion of the class group $h_3(\mathbb Q(\sqrt {-D}))\ll _\varepsilon D^{{1}/{3}+\varepsilon }$ [Reference Ellenberg and Venkatesh7, Proposition 2] into the mechanism as in the proof of [Reference Helfgott and Venkatesh14, Theorem 4.5].

Remark 3.3. Proposition 3.2 gives a better bound than the significantly improved result on the 2-torsion of the class group of Bhargava et al. [Reference Bhargava, Shankar, Taniguchi, Thorne, Tsimerman and Zhao2]. This is because [Reference Bhargava, Shankar, Taniguchi, Thorne, Tsimerman and Zhao2, Theorem 1.2(c)] gives the bound on general elliptic curves rather than Mordell curves. It would be interesting to know if a bound on the $2$ -torsion can be exploited to improve upon Conjecture 3.1.

Remark 3.4. It is possible to prove a weak version of Conjecture 3.1 for special D. The weak version means that for each $\varepsilon>0$ and a fixed integer $D\not =0$ ,

(3.1) $$ \begin{align} \#\{(x,y)\in\mathbb Z^2: |x|,|y|\leqslant B, y^2=x^3+D\}\ll_{\varepsilon} (B|D|)^{\varepsilon}. \end{align} $$

For any D, this weak version follows from Heath-Brown’s result [Reference Heath-Brown, Perelli and Viola11, Theorem 2]. We take $G(X,Y,Z)=ZY^2-X^3-DZ^3$ in Heath-Brown’s theorem. This cubic form covers all solutions on our Mordell curve and leads to the estimate in (3.1). Heath-Brown assumed the Birch–Swinnerton-Dyer conjecture and the Riemann hypothesis for L-functions of elliptic curves to obtain his result. Under these hypotheses, this version suffices to prove the second assertion in Theorem 1.2. This is because we may take $B=N$ in the weak version above and use the bound as in the proof below.

Proof of Theorem 1.2.

As in (2.2), we consider a majorant $g(x)$ of $f(x)$ , say $g(x)=ax^2+b$ with $b\not =0$ . Then, we find an upper bound for $\mathcal {M}(E,D)$ by fixing $e\in [E,2E]$ and count the number of solutions $(d,n)$ to $an^2=e^2d^3-b$ . By Proposition 3.2 with $(x,y)=(e^2ad, e^2a^2n)$ and $D=-e^4ba^3$ , we obtain $\mathcal {M}(E,D)\ll _{\varepsilon , f} E\cdot |E|^{4\varpi _0+\varepsilon }=E^{1+4\varpi _0+\varepsilon }$ . Thus, we obtain the total bound

$$ \begin{align*}\ll_{\varepsilon, f} N^\varepsilon\bigg((\log N)\sup_{\substack{E\leqslant N\\ E^2D^3\asymp N^2}}\min(D,E^{1+4\varpi_0})\bigg)\ll N^{\varpi+\varepsilon},\end{align*} $$

where $\varpi ={2(1+4\varpi _0)}/{(5+12\varpi _0)}$ for the optimal choice $E=N^{{2}/{(5+12\varpi _0)}}$ . The second assertion can be proved similarly by replacing $\varpi _0$ by $0$ . The theorem follows.

Remark 3.5. It seems difficult to tackle this problem with our naive method using the mechanisms of Heath-Brown [Reference Heath-Brown, Perelli and Viola11] or Salberger (as mentioned in [Reference Browning4]). We estimate the number of points $(d,n)$ that satisfy $q(d)-p(n)-b:=e^2d^3-an^2-b=0$ , where $b\not =0$ , $p(x)$ is a polynomial of degree $2$ and $q(x)$ is a polynomial of degree $3$ . Then, we proceed as in Reuss’s work [Reference Reuss15] and obtain via [Reference Heath-Brown, Perelli and Viola11, Theorem 15] that the points $(d,n)$ lie on at most $N^\eta \sqrt {D}$ auxiliary curves. Thus, using Bézout’s theorem, the number of points considered is $\ll _\eta N^\eta \sqrt {D}$ . Upon invoking Estermann’s bound in the first sum, we obtain the upper bound

$$ \begin{align*}\ll_\eta N^\eta\bigg((\log N)\sup_{\substack{E\leqslant N\\ E^2D^3\asymp N^2}}\min(D,E\sqrt{D})\bigg)\ll N^{{1}/{2}+\eta},\end{align*} $$

where the supremum occurs at $E=N^{1/4}$ . This is, however, weaker than what we expected in Theorem 1.2.

3.2 Conditional result based on the $abc$ conjecture

In this subsection, we show that the number $\varpi $ in Theorems 1.1 and 1.2 can be taken arbitrarily close to $0$ under the $abc$ conjecture (Conjecture 3.6 below).

For a positive integer n, we write $\text {rad}(n)=\prod _{p|n} p$ .

Conjecture 3.6 ( $abc$ conjecture).

For any fixed $\varepsilon>0$ , if $a+b=c$ for pairwise coprime positive integers $a,b,c$ , then $c<K_\varepsilon \cdot \text {rad}(abc)^{1+\varepsilon }$ for some constant $K_\varepsilon $ .

Proposition 3.7. Let $\varepsilon>0$ and $f(x)$ be an admissible quadratic polynomial. Under the $abc$ conjecture, the number of $n\leqslant N$ for which $f(n)$ is square-full is at most $O_{\varepsilon , f}(N^\varepsilon )$ .

Proof. Assume that the $abc$ conjecture holds. We focus on counting the solutions $(n,e,d)$ in $ce^2d^3=n^2+b$ for fixed constants $c,b$ in (1.1). Here, we remove the factor a from both p and q in (1.1), and the remaining denominator $4a$ is moved and relabelled as c. The sizes of c and b are dependent on f. Then, we determine the condition on d to satisfy $ce^2d^3=n^2+b$ . For $b>0$ and $\gcd (b,n)=\ell $ , we transform the equation into

$$ \begin{align*} \frac{ce^2d^3}{\ell\ell_1} = \ell' n^{\prime 2}+b", \end{align*} $$

where $n=\ell n'$ and $b=\ell b'$ with $\gcd (n',b')=1$ . We also let $\ell _1=\gcd (\ell ,b')$ with $b'=\ell _1 b", \ell =\ell _1\ell '$ . Thus, there exists an absolute constant $C_\varepsilon $ such that

$$ \begin{align*} \frac{ce^2d^3}{\ell\ell_1}&\leqslant C_\varepsilon\text{rad}\bigg(\frac{ce^2d^3}{\ell\ell_1}\cdot \ell'n^{\prime 2}\cdot b"\bigg)^{1+\varepsilon} \\ &\leqslant C_\varepsilon \text{rad}\bigg(\frac{ce^2d^3}{\ell_1^2}n'b"\bigg)^{1+\varepsilon}\leqslant C_\varepsilon \text{rad}(ce^2d^3n'b")^{1+\varepsilon}\ll_{\varepsilon, f} e^{1+\varepsilon}d^{1+\varepsilon}n^{\prime 1+\varepsilon}. \end{align*} $$

Hence, $e^{1-\varepsilon }d^{2-\varepsilon }\ll _{\varepsilon , f} n^{1+\varepsilon },$ and since $e^2d^3\asymp _{f} n^2$ , we obtain $d\ll _{\varepsilon , f} N^\varepsilon $ for arbitrarily small $\varepsilon>0$ . Estermann’s bound and the number of possibilities of d imply the claim. For $b<0$ , the argument is similar.

Remark 3.8. Since certain $f(x)$ produce infinitely many square-full numbers, the bound in Proposition 3.7 is best possible as a power of N.

4 The determinant method and the proof of Theorem 1.1

In this section, we prove Theorem 1.1, following Heath-Brown [Reference Heath-Brown13] and Reuss [Reference Reuss15]. Fixing $\eta>0$ , we determine the solutions $(e,d,n)$ to $e^2d^3=n^2+\alpha ^2$ with d square-free. Recall that we have parameters $E,D,N$ , where $E<e\leqslant 2E, {D<d\leqslant 2D}$ and ${N<n\leqslant 2N}$ . Thus, $E^2D^3\asymp N^2$ since $\alpha \leqslant N^\varpi $ and $\varpi ={29}/{100}$ . We have ${e=x_1^2+x_2^2}$ and $d=y_1^2+y_2^2$ by the sum of two squares argument, and d is square-free. By the unique factorisation property of $\mathbb Z[i]$ , $(x_1+ix_2)^2(y_1+iy_2)^3=n+\alpha i$ and, after taking the imaginary part,

(4.1) $$ \begin{align} (x_1^2-x_2^2)(3y_1^2y_2-y_2^3)+2x_1x_2(y_1^3-3y_1y_2^2)=\alpha. \end{align} $$

We shall assume that $|x_1|\leqslant |x_2|$ and $x_1,x_2$ have the same signs. If $|x_1|\geqslant |x_2|$ , we may change the roles of $x_1$ and $x_2$ , and change the sign of $y_2$ . If $x_1,x_2$ have different signs, we can change the signs of $y_1$ and $x_2$ . Now, we claim that

$$ \begin{align*}\max(|3y_1^2y_2-y_2^3|,|y_1^3-3y_1y_2^2|)\gg D^{{3}/{2}}.\end{align*} $$

The proof will assume $|y_1|\leqslant |y_2|$ ; by symmetry, the other case can be handled similarly. For the case $|y_2|\cdot |3y_1^2-y_2^2|\geqslant |y_1|\cdot |y_1^2-3y_2^2|$ , we first suppose that $y_2^2<3y_1^2$ . Then, $|y_2|(3y_1^2-y_2^2)\geqslant |y_1||y_1^2-3y_2^2|\geqslant |y_1|(3y_2^2-y_1^2)$ . Since d is square-free, we see that $|y_1|\not =|y_2|$ , for otherwise, $d=2y_1^2$ is not square-free and $d\gg D$ is large for $N\gg _\eta 1$ , so that $3t^2-1\geqslant t(3-t^2)$ for $1\geqslant t:=|y_1/y_2|$ . Thus, $0\leqslant t^3+3t^2-3t-1\leqslant 3t^2-3t=3t(t-1)$ and $t=1$ so $|y_1|=|y_2|$ , which is not possible. Therefore, $y_2^2\geqslant 3y_1^2$ and, with the same setup, we obtain $0\leqslant t^3-3t^2-3t+1=(t-1)^3-6t+2\leqslant 2-6t$ and $|y_1|\leqslant \tfrac 13|y_2|$ . The maximum in this case is $\gg |y_2|(y_2^2-3y_1^2)\gg D^{3/2}$ . Similarly, for the case $|y_2||3y_1^2-y_2^2|<|y_1||y_1^2-3y_2^2|$ , with the same notation, we have either

$$ \begin{align*}t(3-t^2)>3t^2-1\quad\text{or}\quad t(3-t^2)>1-3t^2,\end{align*} $$

and the second inequality is impossible, as it is equivalent to $(t+1)(t^2-4t+1)<0$ . The first corresponds to $(t-1)(t^2+4t+1)<0$ , which is always true, and corresponds to $t^2<3$ , which implies $y_2^2<3y_1^2$ . Again, the maximum of the two terms is $\gg D^{3/2}$ .

On taking $q_1(y_1,y_2)$ to be the larger of the two, which has the property that its absolute value is $\gg D^{3/2}$ , and taking $z_1,z_2$ appropriately to be polynomials in $x_1,x_2$ , we obtain $q_1(y_1,y_2)z_1+q_2(y_1,y_2)z_2=\alpha $ .

We may assume that $D\geqslant N^\varpi $ . Estermann [Reference Estermann9, (16)] provides a bound on

$$ \begin{align*}\#\{(n,e)\in\mathbb Z^2:|n|,|e|\leqslant N, n^2=d^3e^2-\alpha^2\}\ll \log (N+|d\alpha|).\end{align*} $$

For $D<N^\varpi $ , the contribution is

$$ \begin{align*} \sum_{\substack{N<n\leqslant 2N\\ e^2d^3=n^2+\alpha^2\\ D<d\leqslant 2D}}\mu^2(d)\leqslant\sum_{d\leqslant 2N^\varpi} \sum_{\substack{n^2=d^3e^2-\alpha^2\\|n|,|e|\leqslant 2N}} 1\ll N^\varpi\log N, \end{align*} $$

which is acceptable. By the triangle inequality,

$$ \begin{align*}D^{{3}/{2}}|z_1|\ll |q_1(y_1,y_2)||z_1|\leqslant\alpha +|q_2(y_1,y_2)||z_2|\ll D^{{3}/{2}}|z_2|,\end{align*} $$

since $\alpha \leqslant N^\varpi \leqslant D$ and here, we assume that $|z_1z_2|\not = 0$ . For if $|z_1z_2|=0$ , then $x_1=x_2$ , and (4.1) becomes $2x_1^2y_1(y_1^2-3y_2^2)=\alpha $ . The number of $(x_1,x_2,y_1,y_2)$ satisfying such an equation is $\ll \tau (\alpha )^{10}\ll N^\varepsilon $ . Thus, $|z_1|\ll |z_2|$ and $|z_2|\gg \max (|z_1|,|z_2|)\gg E$ . Hence,

(4.2) $$ \begin{align} t=-\dfrac{q_2(s,1)}{q_1(s,1)}+O(E^{-1}D^{-{3}/{2}})=-\dfrac{q_2(s,1)}{q_1(s,1)}+O(N^{-1})=:\phi(s)+O(N^{-1}), \end{align} $$

where $s=y_1/y_2$ and $t=z_1/z_2$ if $|y_1|\leqslant |y_2|$ and $s=y_2/y_1$ in the other case, which can be handled in a similar way. So we may assume $s,t\ll 1$ . Therefore, the problem reduces to counting points $(s,t)$ that lie close to the curve $t=\phi (s)$ . We write $w=x_1/x_2$ which is positive since $x_1,x_2$ have the same signs and $w\ll 1$ .

4.1 The determinant method

In this subsection, we construct another equation in a bi-homogeneous form of the same variables as in (4.2) for each interval that we split. The bound is plausible since we have the existing bound for counting points on $G(x_1,x_2;y_1,y_2)=0$ , for an absolutely irreducible bi-homogeneous polynomial $G(x_1,x_2;y_1,y_2)\in \mathbb Z[x_1,x_2,y_1,y_2]$ .

Our plan is as follows. We choose a parameter $M\leqslant N$ , which will be specified later in terms of $E,D$ . Then, we divide the intervals of possible values of $s\ll 1$ into $O(M)$ intervals of the form $I=(s_0, s_0+M^{-1}]$ . By Taylor’s theorem, if we write $s=s_0+u$ with $u=O(M^{-1})$ , then $\phi (s)=\phi (s_0)+P(u)+O(N^{-1})$ , where $P(u)$ is a polynomial in u without a constant term and all coefficients of size $O(1)$ . Thus, $|P(u)|\ll M^{-1}$ . Write $\tau (w):=t=\phi (s)+O(N^{-1})$ , where $0<w=x_1/x_2\ll 1$ from our conditions. If $\tau (w)=(w^2-1)/2w$ , then we obtain

$$ \begin{align*}\dfrac{w^2-1}{2w}=\phi(s_0)+P(u)+O(N^{-1}).\end{align*} $$

Since $w>0$ and $P(u)$ is small,

$$ \begin{align*}w=\phi(s_0)+P(u)+\sqrt{(\phi(s_0)+P(u))^2+1+O(N^{-1})}.\end{align*} $$

By Taylor’s theorem, $\sqrt {s+c}=\sqrt {c}+(2\sqrt {c})^{-1}s+O(s^2)$ for $s\in \mathbb R$ around zero and a constant $c\geqslant 1$ . Hence,

$$ \begin{align*} w &= \phi(s_0)+P(u)+\sqrt{1+(\phi(s_0)+P(u))^2}+\dfrac{O(N^{-1})}{2\sqrt{1+(\phi(s_0)+P(u))^2}} \\ &= c_{s_0}+Q(u)+O(N^{-1}), \end{align*} $$

using Taylor’s theorem for the square root, and the equation holds for some polynomial Q in u. Similarly, if $\tau (w)=2w/(w^2-1)$ , we obtain

$$ \begin{align*} w=\dfrac{1+\sqrt{1+(\phi(s_0)+P(u)+O(N^{-1}))^2}}{\phi(s_0)+P(u)}\ll 1, \end{align*} $$

and as $P(u)$ is small, $|\phi (s_0)|\gg 1$ . By Taylor’s theorem applied to $\sqrt {1+(c+s)^2}$ for $s\in \mathbb R$ around zero and a constant $c\geqslant 1$ ,

$$ \begin{align*} w = \dfrac{1+\sqrt{1+(\phi(s_0)+P(u))^2}}{\phi(s_0)+P(u)}+\dfrac{O(N^{-1})}{\sqrt{1+(\phi(s_0)+P(u))^2}} =c_{s_0}+Q(u)+O(N^{-1}), \end{align*} $$

since $|\phi (s_0)|\gg 1$ and $P(u)$ is small. Since $P(u)\ll M^{-1}$ and M will be chosen so that $\log M\asymp \log N$ , finitely many terms are accumulated in $Q(u)$ with the residual term of order $O(N^{-1})$ . Therefore, in either case, w can be represented by a polynomial $Q(u)$ such that $Q(u)\ll M^{-1}$ with error $O(N^{-1})$ .

Suppose that there are J solutions to (4.2) in the interval I. Upon introducing new parameters $K,L$ that will be dependent merely on $\eta $ , we construct a $J\times H$ matrix

$$ \begin{align*}\mathfrak M= \begin{pmatrix} 1 & s_1 & w_1 & \cdots & s_1^k w_1^\ell & \cdots & s_1^K w_1^L\\ 1 & s_2 & w_2 & \cdots & s_2^k w_2^\ell & \cdots & s_2^K w_2^L\\ \vdots & & & \vdots & & & \vdots \\ 1 & s_J & w_J & \cdots & s_J^k w_J^\ell & \cdots & s_J^K w_J^L\\ \end{pmatrix},\end{align*} $$

where $H=(K+1)(L+1)$ . If $\text {rank}(\mathfrak M)<H$ , there is a nonzero vector $\mathbf {c}$ such that

$$ \begin{align*}\mathfrak M\mathbf{c}=\mathbf{0}.\end{align*} $$

With this nonzero vector $\mathbf {c}$ , we can consider the polynomial $C_I(s,w)=\sum _{h=1}^H c_hm_h(s,w)$ , where we use the label $m_h(s,w)$ to run through the monomials $s^kw^\ell $ appearing in each row of $\mathfrak {M}$ for $k\leqslant K$ and $\ell \leqslant L$ so that $C_I(s_j, w_j)=0$ for all $j\leqslant J$ . By reducing the matrix $\mathfrak {M}$ to row-reduced echelon form, we see that $\mathbf {c}\in \mathbb Q^H$ has rational entries and we may clear the denominator of $C_I$ , which is $\ll N^A$ for some fixed A. Then, the coefficients are integral of size $\ll N^{\kappa (K,L)}$ . The fact that $\text {rank}(\mathfrak {M})$ is strictly less than H follows from

(4.3) $$ \begin{align} \bigg(\prod_{j\leqslant H} y^K_{2,j}x^L_{2,j}\bigg)\Delta\in\mathbb Z, \end{align} $$

where $\Delta $ is the subdeterminant of the first $H\times H$ matrix, and we assume that $J\geqslant H$ . We can choose M appropriately so that $\Delta =0$ . It suffices to prove, for an appropriate M, that $|\Delta |\leqslant C_{K,L} D^{-{KH}/{2}}E^{-{LH}/{2}}$ for a sufficiently small constant $C_{K,L}$ , which is chosen so that the absolute value of the term in (4.3) is smaller than $1$ . This forces $\Delta $ to be $0$ .

Following the proof of Heath-Brown [Reference Heath-Brown13], for $V\asymp N$ , the Taylor expansion $w=c_{s_0}+P(u)+v$ has $v\ll V^{-1}$ and $u\ll M^{-1}$ . We now order the values of $M^{-k}V^{-\ell }$ , $k\leqslant K$ , $\ell \leqslant L$ , decreasingly as $1=M_0,M_1,\ldots .$ From Heath-Brown [Reference Heath-Brown12, Lemma 3],

$$ \begin{align*}\Delta\ll_H \prod_{h\leqslant H} M_h.\end{align*} $$

Let us denote $M_H=W^{-1}$ . Then, $M^{-k}V^{-\ell }\geqslant M_H$ if and only if $k\log M+\ell \log V\leqslant \log W$ . The number of pairs $k,\ell $ satisfying this inequality is

$$ \begin{align*}\dfrac{\log^2 W}{2\log M\log V}+O\bigg(\dfrac{\log W}{\log N}\bigg)+O(1),\end{align*} $$

which equals H. Since $\log M,\log V\ll \log N$ ,

$$ \begin{align*}\log W=\sqrt{2H\log M\log V}+O(\log N).\end{align*} $$

Moreover,

$$ \begin{align*} \log\prod_{h\leqslant H} M_h &=\sum_{\substack{k.\ell \\ k\log M+\ell \log V\leqslant \log W}} -(k\log M+\ell\log V) \\ &= -\dfrac{\log^3 W}{3\log M\log V}+O\bigg(\dfrac{\log^2 W}{\log N}\bigg) \\ &=-H^{{3}/{2}}\cdot\dfrac{2\sqrt{2}}{3}\sqrt{\log M\log V} +O(H\log N), \end{align*} $$

by using the estimate for $\log W$ above. This gives

$$ \begin{align*}\log |\Delta|\leqslant O_H(1)-H^{{3}/{2}}\cdot\dfrac{2\sqrt{2}}{3}\sqrt{\log M\log V}+O(H\log N).\end{align*} $$

Thus, $\Delta =0$ if

$$ \begin{align*} \dfrac{K}{2}\log D+\dfrac{L}{2}\log E\leqslant (KL)^{{1}/{2}}\dfrac{2\sqrt{2}}{3}\sqrt{\log M\log V}+O_{K,L}(1)+O(\log N). \end{align*} $$

Taking $K=[L\log E/\kern-1.5pt\log D]$ , the condition becomes

$$ \begin{align*} L\log E\leqslant L\cdot\dfrac{2\sqrt{2}}{3}\sqrt{\log M\log V}\sqrt{\dfrac{\log E}{\log D}}+O_{L}(1)+O(\log N). \end{align*} $$

For small fixed $\delta>0$ , if

$$ \begin{align*}\dfrac{2\sqrt{2}}{3}\sqrt{\log M\log V}\sqrt{\dfrac{\log E}{\log D}}\geqslant (1+\delta)\log E\end{align*} $$

and $L,N\gg _\delta 1$ , we will obtain the desired inequality. Since $V\gg N$ , we have proved the following lemma.

Lemma 4.1. Let $\eta>0$ be given and let $M\in [D, N]$ satisfy

$$ \begin{align*}\log M\geqslant \dfrac{9}{8}(1+\eta)\dfrac{\log E\log D}{\log N}.\end{align*} $$

Then, for any interval $I=[s_0,s_0+M^{-1}]$ , there exists a nonzero polynomial $C_I(s,t)$ with integral coefficients such that all solutions satisfy

$$ \begin{align*}C_I\left(y_1/y_2,x_1/x_2\right)=0,\end{align*} $$

with $y_1/y_2\in I$ . Moreover, $\deg (C_I)=O_\eta (1)$ and its coefficients are $O_\eta (N^{\kappa })$ for some $\kappa =\kappa (\eta )$ .

As in [Reference Heath-Brown13], we may assume without loss of generality that this polynomial $C_I$ is absolutely irreducible. So we can apply the existing bound on counting solutions noted at the beginning of this subsection.

4.2 Counting the solutions and finishing the proof

In the previous subsection, we constructed an auxiliary equation involving an absolutely irreducible bi-homogeneous polynomial $C_I$ in each fixed interval $I{\kern-1pt}={\kern-1pt}(s_0,s_0{\kern-1pt}+{\kern-1pt}M^{-1}]$ . In the interval ${I{\kern-1pt}={\kern-1pt}(s_0,s_0{\kern-1pt}+{\kern-1pt}M^{-1}]}$ ,

$$ \begin{align*}s_0<s=\dfrac{y_1}{y_2}\leqslant s_0+M^{-1}.\end{align*} $$

It follows that $|y_1-s_0y_2|\leqslant D^{1/2}M^{-1}$ because $|y_2|\leqslant D^{1/2}$ . Let

$$ \begin{align*}\Lambda = \{(D^{-1/2}M(y_1-s_0y_2), D^{-1/2}y_2):(y_1,y_2)\in\mathbb Z^2\}.\end{align*} $$

Then, $\Lambda $ is a lattice of determinant $D^{-1}M$ . We are interested in points $(\alpha _1,\alpha _2)\in \Lambda $ falling in the square

$$ \begin{align*}S=\{(\alpha_1,\alpha_2):\max(|\alpha_1|,|\alpha_2|)\leqslant 1\}.\end{align*} $$

Let $\mathbf {g}^{(1)}$ be the shortest nonzero vector in the lattice and $\mathbf {g}^{(2)}$ be the shortest vector not parallel to $\mathbf {g}^{(1)}$ . The two vectors form a basis for $\Lambda $ and $\lambda _1\mathbf {g}^{(1)}+\lambda _2\mathbf {g}^{(2)}\in S$ only when $|\lambda _1|\ll |\mathbf {g}^{(1)}|^{-1}=: L_1$ and $|\lambda _2|\ll |\mathbf {g}^{(2)}|^{-1}=:L_2$ . Thus, $L_1\gg L_2$ and $(L_1L_2)^{-1}=|\mathbf {g}^{(1)}||\mathbf {g}^{(2)}|\ll \det (\Lambda )=D^{-1}M$ , which implies $L_1L_2\gg DM^{-1}$ . Hence, we may replace $(y_1,y_2)$ by $(\lambda _1,\lambda _2)$ . We may argue in exactly the same way to replace $x_1,x_2$ by $\tau _1,\tau _2$ . Now, we have the equations

(4.4) $$ \begin{align} G_0(\lambda_1,\lambda_2;\tau_1,\tau_2) &= \alpha, \end{align} $$

(4.5) $$ \begin{align} G_1(\lambda_1,\lambda_2;\tau_1,\tau_2) &= 0, \end{align} $$

which correspond, respectively, to (4.1) and the vanishing equation from Lemma 4.1. We know that $G_0$ is bi-homogeneous of degree $(3;2)$ . Suppose that a bi-homogeneous polynomial $G_1$ has degree $(a;b)$ . When $a,b\geqslant 2$ , we can get a satisfactory bound from the following lemma derived from Heath-Brown [Reference Heath-Brown13, Lemma 2].

Lemma 4.2. Let $G(x_1,x_2;y_1,y_2)$ be an absolutely irreducible bi-homogeneous polynomial with integer coefficients of degree $(a; b)$ with $a,b\geqslant 1$ . Let $\varepsilon>0$ be given. Then, for $X\geqslant 1$ , there exist $O_{a,b,\varepsilon }(X^{{2}/{b}+\varepsilon }\lVert G\rVert ^\varepsilon )$ points $(a_1,a_2,b_1,b_2)\in \mathbb Z^4$ satisfying the conditions $\gcd (a_1,a_2) = \gcd (b_1,b_2)=1, G(a_1,a_2;b_1,b_2) = 0$ and $\max \{a_1,a_2\}\leqslant X$ .

It is possible to assume that $\gcd (\lambda _1,\lambda _2)=\gcd (\tau _1,\tau _2)=1$ at the expense of a factor of $\alpha ^\varepsilon $ . Suppose that $\gcd (\lambda _1,\lambda _2)=\lambda $ . Then, $\lambda ^3\mid \alpha $ as $G_0$ is homogeneous in $\lambda _1,\lambda _2$ . If this happens with $\lambda>1$ , we split this case and count the solutions from it. There are $O(\tau (\alpha ))=O(\alpha ^{\varepsilon /2})$ possibilities of such equations. Similarly, there exist $O(\alpha ^{\varepsilon /2})$ possible equations when we determine a pair of $\tau _1,\tau _2$ .

We present here an elementary result on counting integral solutions with some conditions on certain bivariate forms.

Lemma 4.3. Let N be a large positive number and $\alpha $ be as in Theorem 1.1. Let $c,d$ be a pair of integers with $|c|,|d|\leqslant N^{1000}$ . Then, there are $O_\varepsilon (N^\varepsilon )$ solutions $(y_1,y_2)\in \mathbb Z^2$ with $|y_1|,|y_2|\leqslant N$ to the equation

$$ \begin{align*} P_{c,d}(y_1,y_2):=c(3y_1^2y_2-y_2^3)+d(y^3_1-3y_1y_2^2)=\alpha. \end{align*} $$

Proof. Without loss of generality, we assume that $\gcd (c,d)=1$ since $\alpha \leqslant N^{{29}/{100}}$ and there are at most $\tau (\alpha )\ll _\varepsilon N^\varepsilon $ possible cases to determine. We first prove that $P(y_1,y_2):=P_{c,d}(y_1,y_2)$ is not a product of linear factors. Suppose that there exists a linear factor $qy_1-py_2$ with $p,q\in \mathbb Z$ and $\gcd (p,q)=1$ . Then,

$$ \begin{align*} dp^3+3cp^2q-3dpq^2-cq^3=0, \end{align*} $$

which implies that $p\mid c$ and $q\mid d$ . Let $c=pc',d=qd'$ with $\gcd (c',d')=1$ . Substituting in the equation, we obtain

$$ \begin{align*} \dfrac{c'+3d'}{3c'+d'}=\frac{p^2}{q^2}, \end{align*} $$

from which, by parity checking, $\gcd (c'+3d',3c'+d')=1,2$ or $4$ . If we restrict ourselves to the cases $\gcd (c'+3d',3c'+d')=1,2$ , we have either $c'+3d'=p^{\prime 2}$ and $3c'+d'=q^{\prime 2}$ or $c'+3d'=2p^{\prime 2}$ and $3c'+d'=2q^{\prime 2}$ as the possible systems of equations in integers. In each case, $p^{\prime 2}-3q^{\prime 2}\equiv 0\pmod {4}$ , which is impossible for $\gcd (p',q')=1$ . In this case, $P(y_1,y_2)$ is irreducible over $\mathbb Q$ and we obtain from the main theorem of Bombieri and Schmidt [Reference Bombieri and Schmidt3] involving Thue’s equation that the bound is $O(3^{1+\sup _{d|\alpha }\omega (d)})$ , where $\omega (d)$ is the number of distinct prime factors of d. We have $3^{\sup _{d|\alpha } \omega (d)}\leqslant 3^{\omega (\alpha )}\leqslant \tau (\alpha )^{{\log 3}/{\log 2}}\ll _\varepsilon N^\varepsilon $ . The total bound is therefore $O_\varepsilon (N^{\varepsilon })$ .

When $\gcd (c'+3d',3c'+d')=4$ , we have

$$ \begin{align*} c = \frac{p(3q^2-p^2)}{2}, \quad d = \frac{q(3p^2-q^2)}{2}. \end{align*} $$

The discriminant of the quadratic polynomial

$$ \begin{align*} \frac{P(y_1,y_2)}{qy_1-py_2}=:ay_1^2+by_1y_2+ey_2^2\in\mathbb Q[y_1,y_2] \end{align*} $$

is $b^2-4ae=3(p^2-q^2)^2$ . This means that it is not possible to factor further into two linear factors over $\mathbb Q$ . If there is no such linear factor, $P(y_1,y_2)$ is irreducible over $\mathbb Q$ . Thus, $P(y_1,y_2)$ is not a product of linear factors.

This leaves the case in which $P(y_1,y_2)$ has a quadratic factor. In this case, we bound the number of solutions of $ay_1^2+by_1y_2+ey_2^2=\beta $ for a certain divisor $\beta $ of $\alpha $ . Since there are at most $\tau (\alpha )\ll _\varepsilon N^{\varepsilon }$ divisors, we may consider the case when $\beta $ produces the maximum number of solutions. The equation can be rearranged as

(4.6) $$ \begin{align} (2ay_1+by_2)^2-(4ae-b^2)y_2^2=4a\beta. \end{align} $$

The number of pairs $(y_1,y_2)$ that satisfy (4.6) is at most the number of pairs $(z_1,z_2)\in \mathbb Z^2$ satisfying $z_1^2-\gamma z_2^2:=z_1^2-(4ae-b^2)y_2^2=4a\beta $ and $|z_1|,|z_2|\ll N^{1001}$ . We obtain the bound $O_\varepsilon (N^\varepsilon )$ in this case, since $|\gamma |\ll N^{2000}$ and $4a\beta \ll N^{1001}$ . Hence, the lemma follows.

Now, let us count the number of solutions to (4.4) and (4.5). When $b\geqslant 2$ , we set $X=L_1$ in Lemma 4.2. In (4.5), Lemma 4.1 implies $\lVert G_1\rVert \ll N^{\kappa }$ , where $\kappa =\kappa (\eta )$ . Then, by Lemma 4.2, the number of solutions to (4.5) is $O_\eta (L_1^{1+\eta }N^\eta )$ . Each of these solutions produces at most one solution in (4.4). Therefore, the total number of solutions that satisfy both equations is $O_\eta (N^\eta L_1^{1+\eta })$ . Similarly, for $a\geqslant 2$ , we obtain the contribution $O_\eta (N^\eta T_1^{1+\eta })$ .

For $b=0$ , (4.5) becomes $\tilde {G}_1(\lambda _1,\lambda _2)=0$ for some homogeneous polynomial $\tilde G_1$ . This corresponds to $O(1)$ pairs of $(\lambda _1, \lambda _2)$ , since $\gcd (\lambda _1,\lambda _2)=1$ . Each such d produces $O(\log N)$ pairs $(n,e)$ satisfying ${N}/{2}\leqslant n<N$ and ${E}/{2}\leqslant e<E$ in the equation $e^2d^3=n^2+\alpha ^2$ from the Estermann bound. The total contribution is $O(\log N)$ . In the case $a=0$ , (4.5) becomes $\tilde {G}_1(\tau _1,\tau _2)=0$ for a certain homogeneous polynomial $\tilde {G}_1$ . The number of possible $(\tau _1,\tau _2)$ is $O(1)$ in this case. We then substitute each $(\tau _1,\tau _2)$ into (4.4) and count the $(y_1,y_2)$ in (4.1) for fixed $(x_1,x_2)$ . By Lemma 4.3, the total contribution is $O_\eta (N^\eta )$ .

Then, on considering the cases $b=1$ and $a=1$ as in the work of Heath-Brown (see the discussion before [Reference Heath-Brown13, Lemma 3]), we obtain the following lemma.

Lemma 4.4. For $\eta>0$ , in an interval I, we have $\mathcal M(E,D)\ll _\eta N^\eta \min (L_1^{1+\eta },T_1^{1+\eta })$ .

Proof of Theorem 1.1.

After fixing a parameter M as in Lemma 4.1, we count the possible intervals by fixing $L\ll N$ and consider each interval $L< L_1\leqslant 2L$ . Then, we sum the total contribution dyadically on L by applying Lemma 4.4. If $(y_1,y_2)$ corresponds to $\mathbf {g}^{(1)}$ , then $L_1(y_1-s_0y_2)\ll M^{-1}\sqrt {D}$ and $L_1y_2\ll \sqrt {D}$ . Fix $\varepsilon>0$ . If $L\gg D^{{1}/{2}+\varepsilon }$ , then $y_2=0$ because $L|y_2|\ll L_1|y_2|\ll \sqrt {D}$ and $y_2\in \mathbb Z$ . Also, $y_1=0$ from the first inequality, which is not possible. Hence, $L_1\ll _\varepsilon D^{{1}/{2}+\varepsilon }$ . Write $s_0=y_3M^{-1}$ in $I=(s_0,s_0+M^{-1}]$ so that $y_3\ll M$ since $s_0\ll 1$ . Therefore, the number of intervals for which $L<L_1\leqslant 2L$ is at most the number of triples $(y_1,y_2,y_3)\in \mathbb Z^3$ with ${\gcd (y_1,y_2)=1}$ for which

$$ \begin{align*}y_2y_3=My_1+O(L^{-1}\sqrt{D}), \quad y_2\ll L^{-1}\sqrt{D} \quad\text{and}\quad y_3\ll M.\end{align*} $$

Recall that $L_1\gg L_2$ and $L_1L_2\gg DM^{-1}$ , so $L\gg D^{{1}/{2}}M^{-{1}/{2}}$ . For each value of $y_1$ , there are $O_\eta (N^{\eta }L^{-1}\sqrt {D})$ pairs $(y_2,y_3)$ . Since $L_1$ is of order L, there are $O_\eta (N^\eta L^{-2}D)$ intervals since $y_1\ll L^{-1}\sqrt {D}$ . From Lemma 4.4, $\mathcal {M}(E,D)\ll _\eta N^{2\eta }L^{-2}D(L^{1+\eta })\ll N^{3\eta }\sqrt {DM}$ since $L\gg D^{{1}/{2}}M^{-{1}/{2}}$ . Analogously, we obtain $\mathcal {M}(E,D)\ll N^{3\eta }\sqrt {ME}$ . Therefore, we conclude that $\mathcal {M}(E,D)\ll N^{3\eta }\sqrt {M}\min (\sqrt {E},\sqrt {D}).$

Let M satisfy the condition in Lemma 4.1, dependent on $E,D$ . By Lemma 4.4, the total contribution in each dyadic range is at most

$$ \begin{align*} {N^{\eta}}&\bigg(\sum_{\ell\ll \log N}\sum_{\substack{{N}/{2^{\ell}}<E\leqslant {N}/{2^{\ell-1}}\\E^2D^3\asymp N^2}} \sqrt{M}\min(\sqrt{E},\sqrt{D})\bigg) \\&\ll_\eta {N^\eta}\bigg((\log N)\sup_{\substack{\psi\in[0,1]}} N^{3\psi(1-\psi)/8+\min({\psi}/{2},{(1-\psi)}/{3})}\bigg)\ll_\eta N^{\varpi+\eta}, \end{align*} $$

where $\varpi ={29}/{100}$ for the optimal choice $\psi =\tfrac 25$ . This completes the proof.

5 Comments

We give some comments on possible unconditional extensions of this work. There are two interesting approaches that seem feasible for further investigation.

(1) The extension to higher-degree polynomials. A similar approach to that used in Theorem 1.1 can be used to show for all $m\geqslant 1$ and $\alpha \in \mathbb N$ that

$$ \begin{align*} \#\{n\leqslant N: n^{2m}+\alpha^2\text{ is square-full}\}\ll_{\varepsilon,\alpha,m} N^{{1}/{2}-{3(m+6)}/{4(2m+3)^2}+\varepsilon} \end{align*} $$

for any fixed $\varepsilon>0$ . When $m>1$ , the bound is worse than in the quadratic case because we are unable to exploit the mth power condition. If we could modify the argument and exploit this condition, it might be possible to show that the upper bound approaches $N^\varepsilon $ as the degree of the polynomial increases. Such a result would be significant since we tend to lose control of the error terms for higher-degree polynomials, similar to the square-free case. In fact, we expect that $f(n)$ assumes square-full values finitely many times for any fixed polynomial $f(x)$ of degree at least $3$ without repeated roots. For general polynomials with no repeated roots, one might need to develop analogous results on the counting of integral points on superelliptic curves, defined by the equation $y^k=f(x)$ , where $k\geqslant 2$ and $\deg (f)\geqslant 3$ .

(2) Reducing $\varpi $ . To reduce $\varpi $ in the quadratic case, particularly in Theorem 1.1, one might focus on the contribution when $e\in [N^{{2}/{5}-\delta }, N^{{2}/{5}+\delta }]$ for some $\delta>0$ as this region determines the general bound. As in Remark 3.5 and the discussion around (2.2), we consider counting the number of $(e,d,n)\in \mathbb N^3$ satisfying $e^2d^3=an^2+b$ for a majorant $g(x)=ax^2+b$ with conditions $E<e\leqslant 2E, D<d\leqslant 2D$ and $N<n\leqslant 2N$ . It suffices to count rationals $s,t$ corresponding to

$$ \begin{align*} t:=\frac{e}{n}=\frac{a^{1/2}}{d^{3/2}}+O_b(N^{-2})=:\phi_a(s)+O_b(N^{-2}), \end{align*} $$

where $s=1/d$ and $t=e/n$ . Then, we can employ the determinant method saving on the number of possible intervals since $e\leqslant N^{{2}/{5}+\delta }$ . However, we might need to modify the argument to exploit the fact that s is not just rational, but is of the form $1/d$ . It would also be interesting if one could push the reduction beyond the $\tfrac 25$ threshold assuming the Mordell curve conjecture.